Cyclotomic component shift and multipartition dominance ideals

By the affine recursion in [F1],$T_{a_k}\cdots T_1{X}_1{T}_1\cdots T_{a_k}=Q^{a_k}{L}_{a_k+1},$ including $a_k=0$. By definition $u_{\mathbf{b}}=u_{\mathbf{a}}(L_{a_k+1}-u_k)$. Since the $L_j$ commute, multiplying the first identity by $u_{\mathbf{a}}$ and substituting $L_{a_k+1}=u_k+(L_{a_k+1}-u_k)$ proves the displayed shift identity.

For a multipartition $\lambda$, let $n_c=\mathrm{\mid }{\lambda }^{(c)}\mathrm{\mid }$, $a_c={\sum }_{d<c}{n}_d$, and $\eta =(n_1,\dots ,n_r)$. Write $d(t)=yc$ with $y\in S_{\eta }$ and $c$ the distinguished component-coset representative, so $t=t^{\lambda }\cdot y\cdot c$. The tableau $t^{\lambda }\cdot y$ is standard: $c$ preserves the order of labels within each component interval, so undoing $c$ preserves every row and column inequality of $t$. Let $k$ be the component containing label $1$ in $t^{\lambda }\cdot c$. The factorization of $c$ is $c=s_{a_k}\cdots s_1{c}_0$, where $c_0$ fixes $1$ and uses only $s_i$ with $i\ge 2$. Because $t^{\lambda }\cdot y$ is standard, $y$ fixes the first label $a_k+1$ of component $k$. The element $u_{\mathbf{a}}$ commutes with $T_y$, and $u_{\mathbf{b}}=u_{\mathbf{a}}(L_{a_k+1}-u_k)$ also commutes with $T_y$: $y$ fixes index $a_k+1$, and its remaining simple generators are distant from $L_{a_k+1}$. Also $T_{c_0}$ commutes with $X_1$ by [F6].

Set $C=T_1\cdots T_{a_k}$ and $D=C^{-1}=T_{a_k}^{-1}\cdots T_1^{-1}$. Multiply the identity of step 1.1 on the right by $D{T}_{c_0}$. The middle factors $C$ cancel with $D$, and $T_{c_0}$ commutes with $X_1$, so$u_{\mathbf{a}}{T}_c{X}_1=u_{\mathbf{a}}{h}_1+u_{\mathbf{b}}{h}_2$ for finite-Hecke elements $h_1,h_2$ (explicitly, $h_1=Q^{a_k}{u}_{k}D{T}_{c_0}$ and $h_2=Q^{a_k}D{T}_{c_0}$). Therefore$m_{s,t}{X}_1=T_{d(s)}^{\ast }{x}_{\lambda }{T}_y(u_{\mathbf{a}}{h}_1+u_{\mathbf{b}}{h}_2)\mathrm{.}$ The first summand is governed by [F4], since $u_{\mathbf{a}}{x}_{\lambda }=m_{\lambda }$.

If $k=1$, then $n_1\ge 1$, so $a_c\ge 1$ for every $c\ge 2$. The vector $\mathbf{b}$ has $b_1=1$ and $b_c=a_c$ for $c\ge 2$, hence $u_{\mathbf{b}}$ contains ${\prod }_{c=1}^r(L_1-u_c)=p(L_1)=0$ by [F2]. Thus the shifted second summand vanishes in this case.

Suppose $k>1$. Define the multicomposition $\nu$ by adding a final row of size $1$ to component $k-1$, reducing the first row of component $k$ by $1$, and leaving all other rows unchanged. Its cumulative component vector is $\mathbf{b}$. Let $l={\lambda }_1^{(k)}$, let $G_1$ permute $a_k+1,\dots ,a_k+l$, and let $G_2$ fix $a_k+1$ and permute the other entries in this interval. The length-additive coset decomposition gives$x_{\lambda }={\sum }_{i=1}^l{T}_{c_i}^{\ast }{x}_{\nu },$ where $c_1,\dots ,c_l$ are the distinguished representatives of $G_2\mathrm{\backslash }{G}_1$ in increasing length. Each tableau $u_i=t^{\nu }\cdot c_i\cdot d(s)$ is row standard: the labels $(a_k+1)d(s)<\cdots <(a_k+l)d(s)$ occur increasingly in the first row of component $k$ of $s$, and $c_i$ moves exactly one of them to the new final row of component $k-1$. Consequently ${\mathrm{comp}}_{u_i}(j)={\mathrm{comp}}_s(j)$ except on those moved entries, where the component changes from $k$ to $k-1$. By the tableau action, $d_{\nu }(u_i)=c_i{d}_{\lambda }(s)$. Here $c_i\in G_1\subseteq S_{\lambda }$ and $d_{\lambda }(s)$ is minimal in $S_{\lambda }{d}_{\lambda }(s)$, so parabolic length additivity gives $\mathrm{\ell }(c_i{d}_{\lambda }(s))=\mathrm{\ell }(c_i)+\mathrm{\ell }(d_{\lambda }(s))$ and $T_{c_i}{T}_{d_{\lambda }(s)}=T_{d_{\nu }(u_i)}$. Since $u_{\mathbf{b}}$ commutes with the row subgroup of $\nu$, $x_{\nu }{u}_{\mathbf{b}}=u_{\mathbf{b}}{x}_{\nu }=m_{\nu }$; therefore$T_{d_{\lambda }(s)}^{\ast }{x}_{\lambda }{u}_{\mathbf{b}}={\sum }_{i=1}^l{T}_{d_{\nu }(u_i)}^{\ast }{u}_{\mathbf{b}}{x}_{\nu }={\sum }_{i=1}^l{m}_{u_i,t^{\nu }}^{\nu }\mathrm{.}$ The shifted summand in step 2.1 is this sum multiplied on the right by the finite element $T_y{h}_2$, so [F4] straightens it to standard terms.

The component sizes of $\nu$ move one box from component $k$ to component $k-1$. Before component $k-1$, cumulative dominance counts agree with $\lambda$; at the final row of component $k-1$ the count increases by one. At component $k$, the previous-component increase is canceled by the decrement of its first row; at later components the total sizes agree again. Sorting the rows of each component into a partition can only increase its row-prefix sums. Thus the partition shape obtained from $\nu$ strictly dominates $\lambda$, and every straightened shape in the shifted summand strictly dominates $\lambda$. For $k=1$ the shifted summand was zero by step 2.2. The unshifted summand has the fixed-first-index finite-generator law by [F4]. Hence multiplication of any standard Murphy element by $X_1$ has only shapes $⊵\lambda$, with same-shape first index unchanged and coefficients independent of that first index.

The cyclotomic relation has unit constant coefficient $(-1{)}^r{\prod }_c{u}_c$. Writing $p(X_1)=X_1^r+c_{r-1}{X}_1^{r-1}+\cdots +c_1{X}_1+c_0$ with $c_0\in R^{\times }$, the relation gives$X_1^{-1}=-c_0^{-1}(X_1^{r-1}+c_{r-1}{X}_1^{r-2}+\cdots +c_2{X}_1+c_1)\mathrm{.}$ Also $T_i^{-1}=Q^{-1}(T_i-(Q-1))$, and each $X_j$ is a word in $X_1,T_1,\dots ,T_{j-1}$ by the affine recursion. Therefore closure under right multiplication by $X_1$ and the $T_i$ implies closure under all algebra generators and their inverses.

For each $\lambda$, the preceding steps and [F4] show that $N^{⊵\lambda }$ is a right ideal; applying $\ast$ gives a left ideal. Since $N^{▹\lambda }$ is the sum of $N^{⊵\mu }$ over the finitely many minimal shapes $\mu ▹\lambda$, it too is a two-sided ideal. This proves the stated component-shift route.